I was studying some properties of the spectrum and in my notes I have written that the function $R: \rho(T) \rightarrow (\lambda -T)^{-1} $ is analytic and if $|\lambda| > ||T||$ it is bounded. After that we use Liouville theorem to say that if a function $f:\mathbb{C} \rightarrow \mathbb{C}$ is analytic and bounded then it has to be costant; and we get a contraddiction. My question is, how to prove that $R$ is bounded on all $\rho(T)$?
Thanks for help!
If the spectrum is empty then $(\lambda -T)^{-1}$ exists for all $\lambda$; For $|\lambda| >\|T\|$ we have $(\lambda -T)^{-1}=\frac 1 {\lambda} \sum (\frac T {\lambda})^{n}$. From this it is clear that $\|(\lambda -T)^{-1}\| \to 0$ as $|\lambda| \to \infty$. Hence $\|(\lambda -T)^{-1}\|$ is bounded. [Of course any continuous function is bounded in the disk $\{\lambda: |\lambda | \leq \|T||\}$].